Rapid Recall
- Find a vector orthogonal to both $\vec u=\left<\pi,e,5\right>$ and $\vec v=\left<\sqrt{2},q,7\right>$
Solution
$\left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\ \pi & e& 5 \\ \sqrt{2}& q & 7\end{array} \right|=\left<7e-5q,-(7\pi-5\sqrt{2}), q\pi-\sqrt{2}e \right>$
- Find the area of the parallelogram with the vectors $\vec u=\left<3,4,0\right>$ and $\vec v=\left<0,0,5\right>$ as sides.
Solution
Note that $\vec u \cdot \vec v = 0$, which means that the parallelogram is a rectangle, so $A = \sqrt{3^2+4^2} \cdot 5 = 25$.
- Convert the integral $\int_0^6 \int_0^y x \, dx \, dy$ into polar coordinates.
Solution
$\int_{\pi/4}^{\pi/2} \int_0^{6\csc\theta} (r\cos\theta) \, r \, dr\,d\theta = \int_{\pi/4}^{\pi/2} \int_0^{6\csc\theta} r^2\cos\theta \, dr\,d\theta$ Note: the $d\theta\,dr$ order is a little more involved.
Group problems
- A wire lies along the curve $C$ parametrized by $\vec r(t) = (t^2+1, 3t, t^3)$ for $-1\leq t\leq 2$.
- Compute $ds$. (Remember - a little distance equals the product of the speed and a little time.)
- Set up integral formulas to find $\bar x$, then $\bar y$, then $\bar z$, for the centroid of $C$.
- Let $P=(3,0,0)$, $Q=(0,2,0)$, and $R=(0,0,1)$.
- Find a vector that is orthogonal to both $\vec{PQ}$ and $\vec {PR}$. Call it $\vec n$.
- Find the area of the triangle $\Delta PQR$.
- Let $S=(x,y,z)$ be any point on the plane $PQR$. What is $\vec {PS}\cdot \vec n$ (you should get a number)? Then compute and expand the dot product, to get an equation of the plane.
- Draw the region described the bounds of each integral. (Use the Mathematica notebook Integration.nb to check your work.)
- $\ds\int_{0}^{3}\int_{0}^{9-x^2}\int_{0}^{3-x}dzdydx$
- $\ds\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{\sqrt{1-x^2}}dydxdz$
- $\ds\int_{0}^{3}\int_{0}^{\pi}\int_{0}^{5}rdzdrd\theta$
- $\ds\int_{-1}^{1}\int_{0}^{1-y^2}\int_{0}^{x}dzdxdy$
- Set up an integral formula to compute each of the following:
- The mass of a disc that lies inside the circle $x^2+y^2=9$ and has density function given by $\delta = x+10$
- The $x$-coordinate of the center of mass (so $\bar x$) of the disc above.
- The $z$-coordinate of the center-of-mass (so $\bar z$) of the solid object in the first octant (all variables positive) that lies under the plane $2x+3y+6z=6$.
- The $y$-coordinate of the center-of-mass (so $\bar y$) of the same object.
- Consider $\int_{0}^{4}\int_{x}^{4}\cos(y^2)dydx$.
- Draw the region described by the bounds.
- Swap the order of the bounds on the integral (use $dxdy$ instead of $dydx$) and then actually compute the integral.
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